\documentclass[a4paper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{ctex}
\usepackage{tikz}
\usepackage{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{fontspec}
\usepackage{amsthm}
\usepackage{xltxtra}
\usepackage{pgfplots}
\usepackage{pgfplotstable}
\usepackage{mflogo}
\usepackage{texnames}
\usepackage{graphicx}
\usepackage{titlesec}

\setmainfont{Times New Roman}
\setCJKmainfont[BoldFont=SimHei,ItalicFont = SimSun]{SimSun}

\newtheorem{definition}{Definition}[section]%定义
\newtheorem{theorem}{Theorem}[section]%定理
\newtheorem{axiom}{Axiom}[section]%公理
\newtheorem{lemma}{Lemma}[section]%引理
\newtheorem{proposition}{Proposition}[section]%命题
\newtheorem{corollary}{Corollary}[section]%推论
\newtheorem{remark}{Remark}[section]%注

\title{\heiti\zihao{2} 习题16.2}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{计算下列积分:}
\subsection{$\iint\limits_{D} x \sin (x y) \mathrm{d} x \mathrm{~d} y, D=[0, \pi] \times[0,1]$}
\textbf{解}\quad
$$
	\begin{aligned}
		\iint_{D}x\sin(xy)\mathrm{d}x\mathrm{d}y & =\int_{0}^{\pi}\mathrm{d}x\int_{0}^{1}x\sin(xy)\mathrm{d}y \\
		                                         & =\int_{0}^{\pi}1-\cos x\mathrm{d}x = \pi
	\end{aligned}
$$

\subsection{$\iint\limits_{D} \cos (x+y) \mathrm{d} x \mathrm{~d} y, D=[0, \pi] \times[0, \pi]$}
\textbf{解}\quad
$$
\begin{aligned}
    \iint\limits_{D} \cos (x+y) \mathrm{d} x \mathrm{~d} y&=\int_{0}^{\pi}\mathrm{d}x\int_{0}^{\pi}\cos(x+y)\mathrm{d}y\\
    &=-2\int_{0}^{\pi}\sin x\mathrm{d}x=-4
\end{aligned}
$$

\section{设函数 $f(x, y)$ 在矩形区域 $D=[a, b] \times[c, d]$ 上有连续的二阶偏导数,计算积分$$\iint\limits_{D} \dfrac{\partial^{2} f}{\partial x \partial y}(x, y) \mathrm{d} x \mathrm{~d} y $$}
\textbf{解}\quad
$$
\begin{aligned}
    \iint\limits_{D} \dfrac{\partial^{2} f}{\partial x \partial y}(x, y) \mathrm{d} x \mathrm{~d} y&=\int_{a}^{b}\mathrm{d}x\int_{c}^{d}\dfrac{\partial^{2} f}{\partial x \partial y}(x, y)\mathrm{d}y\\
    &=\int_{a}^{b}\left(\dfrac{\partial f(x,d)}{\partial x}-\dfrac{\partial f(x,c)}{\partial x}\right)\mathrm{d}x\\
    &=f(b,d)-f(a,d)-f(b,c)+f(a,c)
\end{aligned}
$$

\section{改变下列累次积分的积分次序:}
\subsection{$\int_{0}^{1} \mathrm{~d} x \int_{x}^{2 x} f(x, y) \mathrm{d} y$}
\textbf{解}\quad
\begin{figure}[h]
	\centering
	\includegraphics[scale=0.25]{include_picture/16.2.1.png}
	\caption{3.1}
\end{figure}
$$
\begin{aligned}
    \int_{0}^{1} \mathrm{~d} x \int_{x}^{2 x} f(x, y) \mathrm{d} y&=\int_{0}^{1}\mathrm{d}y\int_{\frac{y}{2}}^{y}f(x,y)\mathrm{d}x+\int_{1}^{2}\mathrm{d}y\int_{\frac{y}{2}}^{1}f(x,y)\mathrm{d}x
\end{aligned}
$$

\subsection{$\int_{0}^{1} \mathrm{~d} y \int_{0}^{2 y} f(x, y) \mathrm{d} x+\int_{1}^{3} \mathrm{~d} y \int_{0}^{3-y} f(x, y) \mathrm{d} x$}
\textbf{解}\quad
\begin{figure}[h]
	\centering
	\includegraphics[scale=0.25]{include_picture/16.2.2.png}
	\caption{3.2}
\end{figure}
$$
\begin{aligned}
    \int_{0}^{1} \mathrm{~d} y \int_{0}^{2 y} f(x, y) \mathrm{d} x+\int_{1}^{3} \mathrm{~d} y \int_{0}^{3-y} f(x, y) \mathrm{d} x=\int_0^2\mathrm{d}x\int_{\frac{x}{2}}^{3-x}f(x,y)\mathrm{d}y 
\end{aligned}
$$

\subsection{$\int_{0}^{2 a} \mathrm{d} x \int_{\sqrt{2 a x-x^{2}}}^{\sqrt{2 a x}} f(x, y) \mathrm{d} y$}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->](-0.2,0)--(5,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,5)node[above]{$y$};
	\draw[domain = 0:2] plot(\x,{sqrt(2*\x)}) node[right]{$y=\sqrt{2ax}$};
	\draw[domain = 0:2] plot(\x,{sqrt(2*\x - \x^2)}) node[right]{$y=\sqrt{2ax-x^2}$};
	\draw(2,-0.5)--(2,4)node[right]{$x=2$};
	\draw(0.2,1)--(2.2,1)node[above]{$y=1$};
\end{tikzpicture}
将其转化为Y型区域进行积分,则有
$$
\begin{aligned}
	\int_{0}^{2 a} \mathrm{d} x \int_{\sqrt{2 a x-x^{2}}}^{\sqrt{2 a x}} f(x, y) \mathrm{d} y&=\int_0^a\mathrm{d}y\int_{a+\sqrt{a^2-y^2}}^{2a}f(x,y)\mathrm{d}x\\
	&+\int_0^a\mathrm{d}y\int_{y^2/2a}^{a-\sqrt{a^2-y^2}}f(x,y)\mathrm{d}x+\int_a^{2a}\mathrm{d}y\int_{y^2/2a}^{2a}f(x,y)\mathrm{d}x
\end{aligned}
$$

\subsection{$\int_{0}^{1} \mathrm{~d} x \int_{0}^{x^{2}} f(x, y) \mathrm{d} y+\int_{1}^{2} \mathrm{~d} x \int_{0}^{\sqrt{1-(x-1)^{2}}} f(x, y) \mathrm{d} y$}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->](-0.2,0)--(3,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,3)node[above]{$y$};
	\draw[domain=0:1]plot(\x,{\x^2});
	\draw[domain=1:2]plot(\x,{sqrt(1-(\x-1)^2)});
\end{tikzpicture}
将其转化为Y型区域进行积分
$$
\begin{aligned}
	\int_{0}^{1} \mathrm{~d} x \int_{0}^{x^{2}} f(x, y) \mathrm{d} y+\int_{1}^{2} \mathrm{~d} x \int_{0}^{\sqrt{1-(x-1)^{2}}} f(x, y) \mathrm{d} y&=\int_0^1\mathrm{d}y\int_{\sqrt{y}}^{1+\sqrt{1-y^2}}f(x,y)\mathrm{d}x
\end{aligned}
$$

\section{计算下列二重积分:}
\subsection{$\iint\limits_{D} x \mathrm{~d} x \mathrm{~d} y, D$ 是由直线 $y=2 x, x=2 y, x+y=3$ 所围成的区域}
\textbf{解}\quad
\begin{figure}[h]
	\centering
	\includegraphics[scale=0.25]{include_picture/16.2.3.png}
	\caption{4.2}
\end{figure}
可将积分区域看作X型区域
$$
\begin{aligned}
	\iint_{D}x\mathrm{d}x\mathrm{d}y&=\int_0^1\mathrm{d}x\int_{x/2}^{2x}x\mathrm{d}y+\int_1^2\mathrm{d}x\int_{x/2}^{3-x}x\mathrm{d}y\\
	&=\dfrac{3}{2}
\end{aligned}
$$

\subsection{$\iint\limits_{D} x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y, D$ 是由抛物线 $y^{2}=2 x$ 和直线 $x=2$ 所围成的区域}
\textbf{解}\quad


\subsection{$\iint\limits_{D}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y, D$ 是 $y=x, y=x+a, y=a$ 和 $y=3 a(a>0)$ 所围成的区域}
\textbf{解}\quad
\begin{figure}[h]
	\centering
	\includegraphics[scale=0.25]{include_picture/16.2.4.png}
	\caption{4.3}
\end{figure}
其为Y型区域,从而
$$
\begin{aligned}
	\iint\limits_{D} x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y&=\int_a^{3a}\mathrm{d}y\int_{y-a}^{y}x^2\mathrm{d}x+\int_a^{3a}\mathrm{d}y\int_{y-a}^{y}y^2\mathrm{d}x\\
	&=14a^4
\end{aligned}
$$

\subsection{$\iint\limits_{D} \cos (x+y) \mathrm{d} x \mathrm{~d} y, D=\{(x, y) \| x|+| y \mid \leqslant 1\}$}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->](-0.2,0)--(3,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,3)node[above]{$y$};
	\draw(0,1)--(1,0);
	\draw(1,0)--(0,-1);
	\draw(0,-1)--(-1,0);
	\draw(-1,0)--(0,1);
\end{tikzpicture}
\par
$$
\begin{aligned}
	\iint\limits_{D} \cos (x+y) \mathrm{d} x \mathrm{~d} y&=\int_0^1\mathrm{d}x\int_{x-1}^{1-x}\cos(x+y)\mathrm{d}y+\int_{-1}^0\mathrm{d}x\int_{-x-1}^{1+x}\cos(x+y)\mathrm{d}y\\
	&=\int_{0}^1\left(\sin 1-\sin (2x-1)\right)\mathrm{d}x+\int_{-1}^{0}\left(\sin(2x+1)+\sin 1\right)\mathrm{d}x\\
	&=2\sin 1
\end{aligned}
$$


\subsection{$\iint\limits_{D}|x y| \mathrm{d} x \mathrm{~d} y, D=\left\{(x, y) \mid x^{2}+y^{2} \leqslant R^{2}, R>0\right\}$}
\textbf{解}\quad
\begin{tikzpicture}
	\filldraw[fill = blue!20](0,0) circle(1);
	\draw[->](-0.2,0)--(3,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,3)node[above]{$y$};
\end{tikzpicture}

令$D_1$为$D$在第一象限的部分,则可由函数对称性得
$$
\begin{aligned}
	\iint\limits_{D}|x y| \mathrm{d} x \mathrm{~d} y&=4\iint\limits_{D_1}|x y| \mathrm{d} x \mathrm{~d} y\\
	&=4\int_{0}^{\pi/2}\mathrm{d}\varphi\int_0^R\rho^3\cos\varphi\sin\varphi\mathrm{d}\rho\\
	&=\dfrac{1}{2}R^4
\end{aligned}
$$

\subsection{$\iint\limits_{D}\left[y+x f\left(x^{2}+y^{2}\right)\right] \mathrm{d} x \mathrm{~d} y, D$ 是 $y=x^{2}$ 和 $y=1$ 所围成的区域}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->](-0.2,0)--(3,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,3)node[above]{$y$};
	\draw[domain=-1:1]plot(\x,{abs(\x^2)});
	\draw[domain=-1:1]plot(\x,{1});
\end{tikzpicture}

由于区域关于$y$轴对称且$f(x,y)=xf(x^2+y^2)=-f(-x,y)$,故在$D$上此项积分为$0$.从而
$$
\begin{aligned}
	\iint\limits_{D}\left[y+x f\left(x^{2}+y^{2}\right)\right] \mathrm{d} x \mathrm{~d} y&=\iint\limits_{D}y \mathrm{d} x \mathrm{~d} y\\
	&=\int_{-1}^{1}\mathrm{d}x\int_{x^2}^{1}y\mathrm{d}y\\
	&=1-\dfrac{1}{5}=\dfrac{4}{5} 
\end{aligned}
$$



\section{设 $f(x)$ 为一元连续函数,证明 $$ \int_{0}^{a} \mathrm{~d} x \int_{0}^{x} f(x) f(y) \mathrm{d} y=\dfrac{1}{2}\left(\int_{0}^{a} f(t) \mathrm{d} t\right)^{2}$$}
\begin{proof}
	$$
		\begin{aligned}
			LHS=\int_{0}^{a} \mathrm{~d} x \int_{0}^{x} f(x) f(y) \mathrm{d} y&=\int_0^a\left(f(x)\int_{0}^{x}f(y)\mathrm{d}y\right)\mathrm{d}x\\
			&=\int_0^a\int_0^xf(y)\mathrm{d}y\mathrm{d}\int_0^xf(y)\mathrm{d}y\\
			&=\left.\dfrac{\left(\int_0^xf(t)\mathrm{d}t\right)^2}{2}\right|_{x=0}^{x=a}\\
			&=\dfrac{1}{2}\left(\int_{0}^{a} f(t) \mathrm{d} t\right)^{2}=RHS
		\end{aligned}
	$$
\end{proof}
\begin{proof}
	$$
		\begin{aligned}
			I=\int_0^a\mathrm{d}x\int_0^xf(x)f(y)\mathrm{d}y&=\int_0^a\mathrm{d}y\int_y^af(x)f(y)\mathrm{d}x\\
			&=\int_0^a\mathrm{d}x\int_x^af(x)f(y)\mathrm{d}y\\
			2I=\int_{0}^{a} \mathrm{~d} x \int_{0}^{x} f(x) f(y) \mathrm{d} y+\int_0^a\mathrm{d}x\int_x^af(x)f(y)\mathrm{d}y&=\int_0^a\mathrm{d}x\int_0^af(x)f(y)\mathrm{d}y\\
			&=\left(\int_0^af(y)\mathrm{d}y\right)^2
		\end{aligned}
	$$
\end{proof}

\section{将下列区域上的二重积分 $\iint\limits_{D} f(x, y) \mathrm{d} x \mathrm{~d} y$ 转化为极坐标系下的累次积分}
\subsection{$D=\left\{(x, y) \mid x^{2}+y^{2} \leqslant a^{2}, y>0\right\}$}
\textbf{解}\quad
$$
\begin{aligned}
	\iint\limits_{D} f(x, y) \mathrm{d} x \mathrm{~d} y&=\int_0^{\pi}\mathrm{d}\varphi\int_0^af\cdot\rho\mathrm{d}\rho
\end{aligned}
$$

\subsection{$D=\left\{(x, y) \mid x^{2}+y^{2} \leqslant a y, a>0\right\}$}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->](-0.2,0)--(2,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,2)node[above]{$y$};
	\draw[domain=-0.5:0.5]plot(\x,{(1+sqrt(1-4*\x^2))/2});
	\draw[domain=-0.5:0.5]plot(\x,{(1-sqrt(1-4*\x^2))/2});
\end{tikzpicture}

可以解出$\sin\varphi>0$,从而$\varphi\in[0,\pi]$.
$$
\begin{aligned}
	\iint\limits_{D} f(x, y) \mathrm{d} x \mathrm{~d} y&=\int_0^{\pi}\mathrm{d}\varphi\int_0^{a\sin t}f\cdot\rho\mathrm{d}\rho
\end{aligned}
$$

\subsection{$D=\{(x, y) \mid 0 \leqslant x \leqslant a, 0 \leqslant y \leqslant a\}$}
分成两部分:
$$
\begin{aligned}
	\iint\limits_{D} f(x, y) \mathrm{d} x \mathrm{~d} y&=\int_0^{\pi/4}\mathrm{d}\varphi\int_0^{a/\cos\varphi}\rho\mathrm{d}\rho+\int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{a/\sin\varphi}\rho\mathrm{d}\rho
\end{aligned}
$$


\section{计算下列二重积分}
\subsection{$\iint\limits_{x^2+y^2\leqslant x+y} \sqrt{x^{2}+y^{2}} \mathrm{~d} x \mathrm{~d} y$}
\textbf{解}\quad
极坐标变换,可得$\rho^2\leqslant\rho(\cos\varphi+\sin\varphi)$,从而$\sin\left(\dfrac{\pi}{4}+\varphi\right)\geqslant 0$.
$$
\begin{aligned}
	\iint\limits_{x^2+y^2\leqslant x+y} \sqrt{x^{2}+y^{2}} \mathrm{~d} x \mathrm{~d} y&=\int_{-\pi/4}^{3\pi/4}\mathrm{d}\varphi\int_0^{\cos\varphi+\sin\varphi}\rho^{2}\mathrm{d}\rho\\
	&=\dfrac{2\sqrt{2}}{3}\int_{-\pi/4}^{3\pi/4}\sin^3\left(\varphi+\dfrac{\pi}{4}\right)\mathrm{d}\varphi\\
	&=\dfrac{2\sqrt{2}}{3}\int_0^\pi\sin^3u\mathrm{d}u\\
	&=\dfrac{4\sqrt{2}}{3}\int_0^{\pi/2}\sin^3u\mathrm{d}u\\
	&=\dfrac{8\sqrt{2}}{9}(\text{Stirling})
\end{aligned}
$$

\subsection{$\iint\limits_{1 \leqslant x^{2}+y^{2} \leqslant 4} \dfrac{\sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} x \mathrm{~d} y$}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->](-0.2,0)--(3,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,3)node[above]{$y$};
	\draw(0,0)circle(0.5);
	\draw(0,0)circle(2);
	\draw(0,2)node[above]{$r=4$};
	\draw(0,0.5)node[above]{$r=1$};
\end{tikzpicture}

$$
\begin{aligned}
	\iint\limits_{1 \leqslant x^{2}+y^{2} \leqslant 4} \dfrac{\sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} x \mathrm{~d} y&=\int_{0}^{2\pi}\mathrm{d}\varphi\int_1^4\sin(\rho\pi)\mathrm{d}\rho\\
	&=2\pi\cdot\left(-\dfrac{2}{\pi}\right)\\
	&=-4
\end{aligned}
$$

\subsection{$\iint\limits_{x^2+y^2\leqslant 2x} \sqrt{4-x^{2}-y^{2}} \mathrm{~d} x \mathrm{~d} y $}
\textbf{解}\quad
$\rho^2\leqslant 2\rho\cos\varphi$,可知$2\cos\varphi\geqslant 0$.
$$
\begin{aligned}
	\iint\limits_{x^2+y^2\leqslant 2x} \sqrt{4-x^{2}-y^{2}} \mathrm{~d} x \mathrm{~d} y&=\int_{-\pi/2}^{\pi/2}\mathrm{d}\varphi\int_0^{2\cos\varphi}\rho\sqrt{4-\rho^2}\mathrm{d}\rho\\
	&=\dfrac{8}{3}\int_{-\pi/2}^{\pi/2}1-\sin^3\varphi\mathrm{d}\varphi=\dfrac{8\pi}{3}
\end{aligned}
$$

\subsection{$\iint\limits_{\left(x^{2}+y^{2}\right)^{2} \leqslant 2 a\left(x^{2}-y^{2}\right), a>0}(x+y)^{2} \mathrm{d} x \mathrm{d} y$}
\textbf{解}\quad
\begin{figure}
	\centering
	\includegraphics[scale=0.25]{include_picture/16.2.5.png}
	\caption{7.4}
\end{figure}
$\rho^4\leqslant 2a\rho^2\cos 2\varphi$,显然有$\varphi\in\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]\cup\left[\dfrac{3\pi}{4},\dfrac{5\pi}{4}\right]$.又因为区域为双扭线,从而因其对称性,可将被积函数中的$xy$项消去,从而被积函数化为$x^2+y^2$.
$$
\begin{aligned}
	\iint\limits_{\left(x^{2}+y^{2}\right)^{2} \leqslant 2 a\left(x^{2}-y^{2}\right), a>0}x^2+y^{2} \mathrm{d} x \mathrm{d} y&=\int_{-\pi/4}^{\pi/4}\mathrm{d}\varphi\int_{0}^{\sqrt{2a\cos 2\varphi}}\rho^3\mathrm{d}\rho+\int_{3\pi/4}^{5\pi/4}\mathrm{d}\varphi\int_{0}^{\sqrt{2a\cos 2\varphi}}\rho^3\mathrm{d}\rho\\
	&=\int_{-\pi/4}^{\pi/4}\dfrac{4a^2\cos^22\varphi}{4}\mathrm{d}\varphi+\int_{3\pi/4}^{5\pi/4}\dfrac{4a^2\cos^22\varphi}{4}\mathrm{d}\varphi\\
	&=\dfrac{a^2\pi}{2}
\end{aligned}
$$


\subsection{$\iint\limits_{D} \sqrt{1-\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}} \mathrm{~d} x \mathrm{~d} y, D=\left\{(x, y) \mid \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} \leqslant 1, x \geqslant y \geqslant 0\right\}$, 其中常数 $a, b>0$.}
\textbf{解}\quad
记$x=a\rho\cos\varphi,y=b\rho\sin\varphi$.$\dfrac{\partial(x,y)}{\partial(u,v)}=abr$.
$$
\begin{aligned}
	\iint\limits_{D} \sqrt{1-\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}} \mathrm{~d} x \mathrm{~d} y&=ab\int_{0}^{\arctan\frac{a}{b}}\mathrm{d}\varphi\int_0^{1}\rho\sqrt{1-\rho^2}\mathrm{d}\rho\\
	&=\dfrac{ab\arctan\frac{a}{b}}{3}
\end{aligned}
$$



\section{作适当的变量代换,计算下列二重积分}
\subsection{$\iint\limits_{D} \mathrm{e}^{\dfrac{y}{x+y}} \mathrm{d} x \mathrm{~d} y, \quad D=\{(x, y) \mid x+y \leqslant 1, x \geqslant 0, y \geqslant 0\}$}
记$u=x+y,v=y$,$\dfrac{\partial(x,y)}{\partial(u,v)}=1$.

\begin{tikzpicture}
	\draw[->](-0.2,0)--(3,0)node[right]{$u$};
	\draw[->](0,-0.2)--(0,3)node[above]{$v$};
	\draw(0,0)--(1,1);
	\draw(1,1)--(1,0);
	\draw(1,0)--(0,0);
\end{tikzpicture}

$$
\begin{aligned}
	\iint\limits_{D} \mathrm{e}^{\dfrac{y}{x+y}} \mathrm{d} x \mathrm{d} y&=\int_0^1\mathrm{d}u\int_0^{u}\mathrm{e}^{v/u}\mathrm{d}v\\
	&=\int_0^1u\mathrm{e}-u\mathrm{d}u\\
	&=\dfrac{\mathrm{e}-1}{2}
\end{aligned}
$$

\subsection{$\iint\limits_{D}(x+y) \sin (x-y) \mathrm{d} x \mathrm{~d} y, D=\{(x, y) \mid 0 \leqslant x+y \leqslant \pi, 0 \leqslant x-y \leqslant \pi\}$}
\textbf{解}\quad
令$u=x+y,v=x-y,\dfrac{\partial(x,y)}{\partial(u,v)}=-\dfrac{1}{2}$.
\begin{tikzpicture}
	\draw[->](-0.2,0)--(3,0)node[right]{$u$};
	\draw[->](0,-0.2)--(0,3)node[above]{$v$};
	\draw(0,0)--(0,pi/2)node[left]{$(0,\pi)$};
	\draw(0,pi/2)--(pi/2,pi/2)node[right]{$(\pi,\pi)$};
	\draw(pi/2,pi/2)--(pi/2,0)node[right]{$(\pi,0)$};
	\draw(pi/2,0)--(0,0);
\end{tikzpicture}
$$
\begin{aligned}
	\iint\limits_{D}(x+y) \sin (x-y) \mathrm{d} x \mathrm{~d} y&=-\dfrac{1}{2}\int_0^\pi u\mathrm{d}u\int_0^\pi\sin v\mathrm{d}v\\
	&=-\dfrac{\pi^2}{2}
\end{aligned}
$$

\subsection{$\iint\limits_{D}(x+y) \mathrm{d} x \mathrm{~d} y, D=\{(x, y) \mid 4 \leqslant x+y \leqslant 12,-\sqrt{2 x} \leqslant y \leqslant \sqrt{2 x}, x \geqslant 0\}$}
\textbf{解}\quad
令$u=x+y,v=y,\dfrac{\partial(x,y)}{\partial(u,v)}=1$.
$$
\begin{aligned}
	\iint\limits_{D}(x+y) \mathrm{d} x \mathrm{~d} y&=\int_4^{12}\mathrm{d}u\int_{-1-\sqrt{2u+1}}^{-1+\sqrt{2u+1}}u\mathrm{d}v\\
	&=\int_4^{12}2u\sqrt{2u+1}\mathrm{d}u\\
	&=2\int_4^{12}u\sqrt{2u+1}\mathrm{d}u\\
	&=\dfrac{1}{2}\int_9^{25}\sqrt{u}(u-1)\mathrm{d}u\\
	&=\dfrac{8156}{15}
\end{aligned}
$$

\section{求由曲面 $z=x^{2}+y^{2}$ 和平面 $z=x+y$ 所围成的立体的体积}
\textbf{解}\quad
\begin{tikzpicture}
	\begin{axis}[title=Volum]
		\addplot3[mesh]{x^2+y^2};
		\addplot3[mesh]{x+y};
	\end{axis}
\end{tikzpicture}

边界方程:$x^2+y^2=x+y$,即$\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)=\dfrac{1}{2}$.

$D:\left\{(x,y)\mid \left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)\leqslant\dfrac{1}{2}\right\}$
极坐标变换,$x=\rho\cos\varphi+\dfrac{1}{2},y=\rho\sin\varphi+\dfrac{1}{2}$.
$$
\begin{aligned}
	\iint\limits_D(x+y)-(x^2+y^2)\mathrm{d}x\mathrm{d}y&=\int_0^{2\pi}\mathrm{d}\varphi\int_0^{\frac{\sqrt{2}}{2}}\rho^2(\cos\varphi+\sin\varphi)+\rho-\rho^3-\rho^2(\cos\varphi+\sin\varphi)-\dfrac{\rho}{2}\mathrm{d}\rho\\
	&=\int_0^{2\pi}\mathrm{d}\varphi\int_0^{\frac{\sqrt{2}}{2}}\dfrac{\rho}{2}-\rho^3\mathrm{d}\rho\\
	&=\dfrac{\pi}{8}
\end{aligned}
$$
\section{求由锥面 $z=\sqrt{x^{2}+y^{2}}$,平面 $z=0$ 和圆柱面 $x^{2}+y^{2}=1$ 所围成的立体的体积}
\textbf{解}
\begin{tikzpicture}
	\begin{axis}
		\addplot3[mesh]{sqrt(x^2+y^2)};
		\addplot3[mesh]{0};
	\end{axis}
\end{tikzpicture}

极坐标变换
$$
\begin{aligned}
	\iint\limits_D\sqrt{x^2+y^2}\mathrm{d}x\mathrm{d}y&=\int_0^{2\pi}\mathrm{d}\varphi\int_0^{1}\rho^2\mathrm{d}\rho\\
	&=\dfrac{2\pi}{3}
\end{aligned}
$$

\section{计算由下列曲线所围成的图形的面积:}
\subsection{$x+y=a, x+y=b, y=\alpha x, y=\beta x(0<a<b, 0<\alpha<\beta)$}
\textbf{解}\quad
令$u=x+y,v=y,\dfrac{\partial (x,y)}{\partial(u,v)}=1$.

\begin{tikzpicture}
	\begin{axis}[xmin=0,xmax=4,ymin=0,ymax=4]
		\addplot[blue]{x};
		\addplot[blue]{2*x};
		\addplot[blue]{1-x};
		\addplot[blue]{2-x};
	\end{axis}
\end{tikzpicture}

$$
\begin{aligned}
	S=\int_a^b\mathrm{d}u\int_{\alpha u/(\alpha+1)}^{\beta u/(\beta+1)}\mathrm{d}v&=\left(\dfrac{\beta}{\beta+1}-\dfrac{\alpha}{\alpha+1}\right)\cdot\dfrac{b^2-a^2}{2}
\end{aligned}
$$

\subsection{$\sqrt[4]{\dfrac{x}{a}}+\sqrt[4]{\dfrac{y}{b}}=1, x=0, y=0$}
\textbf{解}\quad
令$u=\sqrt[4]{\dfrac{x}{a}},v=\sqrt[4]{\dfrac{y}{b}},\dfrac{\partial(x,y)}{\partial(u,v)}=16abu^3v^3$
$$
\begin{aligned}
	S=\int_0^1\mathrm{d}u\int_0^{1-u}16u^3v^3\mathrm{d}v&=4ab\int_0^1(1-u)^4u^3\mathrm{d}u\\
	&=\dfrac{1}{70}ab
\end{aligned}
$$

\subsection{$\left(a_{1} x+b_{1} y+c_{1}\right)^{2}+\left(a_{2} x+b_{2} y+c_{2}\right)^{2}=1$, 式中 $a_{1} b_{2} \neq a_{2} b_{1}$}
\textbf{解}\quad
令$u=a_{1} x+b_{1} y+c_{1},v=a_{2} x+b_{2} y+c_{2}$,则$\dfrac{\partial(x,y)}{\partial(u,v)}=\left|\begin{array}{cc}
	a_1 & b_1\\
	a_2 & b_2
\end{array}\right|^{-1}=\dfrac{1}{a_1b_2-b_1a_2}$.再对$u,v$进行极坐标变换.
$$
\begin{aligned}
	S=\dfrac{1}{a_1b_2-b_1a_2}\iint\limits_\Omega\mathrm{d}u\mathrm{d}v&=\dfrac{1}{a_1b_2-b_1a_2}\int_0^{2\pi}\mathrm{d}\varphi\int_0^1\rho\mathrm{d}\rho\\
	&=\dfrac{\pi}{a_1b_2-b_1a_2}
\end{aligned}
$$
其绝对值$\left|\dfrac{\pi}{a_1b_2-b_1a_2}\right|$即为所求.

\subsection{$\left(x^{2}+y^{2}\right)^{2}=2 a^{2}\left(x^{2}-y^{2}\right),\left(x^{2}+y^{2} \geqslant a^{2}, a>0\right)$}
\textbf{解}\quad
\begin{figure}[h]
	\centering
	\includegraphics[scale=0.25]{include_picture/16.2.6.png}
	\caption{11.4}
\end{figure}
极坐标变换
$$
\begin{aligned}
	S=4\int_0^{\frac{\pi}{6}}\mathrm{d}\varphi\int_a^{\sqrt{2a^2\cos 2\varphi}}\rho\mathrm{d}\rho&=\left(\sqrt{3}-\dfrac{\pi}{3}\right)a^2
\end{aligned}
$$


\section{作适当的变换, 将下列二重积分转化为定积分:}
\subsection{$\iint\limits_{x^{2}+y^{2} \leqslant 1} f\left(\sqrt{x^{2}+y^{2}}\right) \mathrm{d} x \mathrm{~d} y$}
\textbf{解}\quad
极坐标变换
$$
\begin{aligned}
	\iint\limits_{x^{2}+y^{2} \leqslant 1} f\left(\sqrt{x^{2}+y^{2}}\right) \mathrm{d} x \mathrm{~d} y&=\int_0^{2\pi}\mathrm{d}\varphi\int_0^{1}f\cdot\rho\mathrm{d}\rho
\end{aligned}
$$


\subsection{$\iint\limits_{D} f(x y) \mathrm{d} x \mathrm{~d} y$, 其中 $D=\{(x, y) \mid x \leqslant y \leqslant 4 x, 1 \leqslant x y \leqslant 2, x>0, y>0\}$}
\textbf{解}\quad
令$u=xy,v=\dfrac{y}{x},\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac{1}{2v}$.
$$
\iint\limits_{D} f(x y) \mathrm{d} x \mathrm{~d} y=\int_{1^4}\mathrm{d}v\int_1^2f(u)\dfrac{1}{2v}\mathrm{d}u
$$



\end{document}